Partial Fractions

  • #1
Why, when a fraction has repeated linear terms in its denominator e.g. (11x2+14x+5)/[(x+1)2(2x+1)] does it have to be split into three partial fractions, A/(x+1) + B/(x+1)2 + C/(2x+1)?
When my first saw this example, my initial reaction was to split it into A/(x+1)2 +B/(2x+1), but after working through this, I realised my method was wrong. Why doesn't it work? I don't want a worked answer to the example because I already know what it is. I just want a genuine logical reason to why the former works and the latter doesn't.
Thanks :)
 

Answers and Replies

  • #2
mathman
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If you leave out 1/(x+1) then the numerator for 1/(x+1)² will be of the form Ax+B.
 
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  • #3
symbolipoint
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Deep_Thinker97,
Try making a couple of simple examples.
Partial fraction decomposition is a way to find what the original fractions were before they were added (or subtracted).

What must be done to sum 3/(x+1)+8/(x+1)^2 ?

How about sum of 3/(x+1)+(8x+2)/(x+1)^2 ?

Try sum of 3/(x+1)+0/(x+1)^2 ?


You always start with proper fractions.
In summing the original fractions, you need to bring some of them to higher terms in order to sum, so you need to find the lowest common denominator.

When you want to decompose a rational expression (partial fraction decomposition), you do not know in advance what are the coefficients or constants in the numerators of the decomposed parts.
 
  • #4
symbolipoint
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If you leave out 1/(x+1) then the numerator for 1/(x+1)² will be of the form Ax+B.
Yes. One does not know the values of the coefficient and constant term until one has finished the decomposition work.

Excuse my comment there, since what I said was not exactly what should have been.
 
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  • #5
symbolipoint
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If you leave out 1/(x+1) then the numerator for 1/(x+1)² will be of the form Ax+B.
Checking about that required some review or restudy. The denominator (x+1) is linear, but it is used as multiplicity of 2; still it is a LINEAR factor. The numerator used for the partial fraction will be just a single constant. If the denominator factor were QUADRATIC AND NOT FACTORABLE in real numbers, then that numerator would be of the form Ax+B.
 
  • #6
mathman
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Checking about that required some review or restudy. The denominator (x+1) is linear, but it is used as multiplicity of 2; still it is a LINEAR factor. The numerator used for the partial fraction will be just a single constant. If the denominator factor were QUADRATIC AND NOT FACTORABLE in real numbers, then that numerator would be of the form Ax+B.
Nope - why not factorable?.
a/(x+1)+b/(x+1)^2=(ax+(a+b))/(x+1)^2
 
  • #7
symbolipoint
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Nope - why not factorable?.
a/(x+1)+b/(x+1)^2=(ax+(a+b))/(x+1)^2
The denominator factor of x+1 is linear. The original rational expression to decompose would contain a denominator (x+1)(x+1)=(x+1)^2=x^2+2x+1, the right side being a FACTORABLE quadratic.

All I could try now is to take the original poster's rational expression and work it on paper to decompose into its partial fractions.
 
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  • #8
symbolipoint
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A/(x+1) + B/(x+1)2 + C/(2x+1)?

I agree with the second partial fraction expression with the constant, B, because the denominator is from having the linear x+1 with exponent 2.

Multiplying by the denominator of the original expression gives
11x^2+14x+5=A(x+1)(2x+1)+B(2x+1)+c(x+1)^2

Working with the right hand side,
arranging in decreasing terms of x,
(2A+C)x^2+(3A+2B+2C)x+(A+B+C)


My best understanding here is that if denominator had a non-factorable x^2+mx+n, then this needs a numerator like Bx+D for a partial fraction.

Deep_Thinker97 already found his answers.
 

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